## Questions d'entretien

Entretien pour Private Wealth Management

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# flip a fair coin, what is the expected number of times one need to flip to get two consecutive head.

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## Réponses aux questions d'entretien

4 réponse(s)

1

xi-i is a martingale with x0-0 = 0 so we have 2^2+2-E[i] = 0 which gives E[i] = 6

xi-i is a martingale le

0

I have an alternative to Charles's answer from the point of view of conditional expectations. The formula is that E[X] = sums of E[X|A] * P(A) where E[X|A] is the conditional expectation of X given event A. Let's consider the following possibilities. If the first toss gives T, then the expected value of tosses under this condition must be 1 time more than the original expectation. Similarly, if one analyzes the case when the first toss gives H, then two possibilities emerge. Either you have a second toss of H, which gives exactly 2, or you have a second toss of T, which gives a conditioned expectation to be added by two. In a summary we have x = (x+1)*P(T) + (x+2)*P(HT) + 2*P(HH) Solve to get x = 6.

Utilisateur anonyme le

7

The previous answer is incorrect. The correct answer is 6. This can be modeled by a simple markov chain: state 0: [1/2 1/2 0] state 1: [1/2 0 1/2] state 2: [0 0 1] where state 0 is having 0 heads in a row (i.e. back at square 1), state 1 is having 1 heads in a row, and state 2 is having 2 heads in a row a.k.a the final state Since we are solving for expected time to reach state 2, we have to solve the system of equations: t0=1+1/2*t0+1/2*t1 t1=1 + 1/2*t0 + 1/2*t2 t2=0 <--if you are already in state 2 i.e. you already have 2 consecutive heads, then you do not need to flip any additional coins because you are already done. Solving the system gets you t0=6. Alternatively, you could use the equation E(x)=(1-p^2)/(p^2*(1-p)), which also gives you 6 when you plug in p=.5

Charles le

3

the probability that you get heads on any given toss is 0.5, since the flips are independent events, the probability of getting two heads consecutively is (.5)(.5)= 0.25=(1/4) thus you would expect to have to flip four times before you would get two consecutive heads.

David Reynolds le

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